Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
double L[maxn], R[maxn];
double sy, a, b;
int n;
inline double F(double x1, double y1, double x2, double y2, double y) {
  return (x2 - x1) * (y - y1) / (y2 - y1) + x1;
}
inline double Time(double x, double y, double l, double r) {
  double ll = max(a, F(x, y, l, 0, sy)), rr = min(b, F(x, y, r, 0, sy));
  return rr - ll;
}
double t[maxn];
void init() {
  for (int i = 1; i <= n; i++) {
    t[i] = t[i - 1] + (1.0 - sy) * (R[i] - L[i]) / 1.0;
  }
}
int main() {
  scanf("%lf%lf%lf", &sy, &a, &b);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lf%lf", L + i, R + i);
  }
  init();
  int q;
  cin >> q;
  double ALL = b - a;
  while (q--) {
    double x, y;
    double ans = 0;
    int fi, se;
    scanf("%lf%lf", &x, &y);
    int l = 1, r = n;
    if (F(a, sy, L[1], 0, y) >= x) {
      fi = 1;
    } else if (F(a, sy, L[n], 0, y) <= x) {
      fi = n + 1;
    } else {
      while (r - l > 1) {
        int mid = r + l >> 1;
        if (F(a, sy, L[mid], 0, y) >= x)
          r = mid;
        else
          l = mid;
      }
      fi = r;
    }
    l = 1, r = n;
    if (F(b, sy, R[n], 0, y) <= x) {
      se = n;
    } else if (F(b, sy, R[1], 0, y) >= x) {
      se = 0;
    } else {
      while (r - l > 1) {
        int mid = r + l >> 1;
        if (F(b, sy, R[mid], 0, y) >= x)
          r = mid;
        else
          l = mid;
      }
      se = l;
    }
    double sum = 0;
    if (se >= fi) {
      sum += (t[se] - t[fi - 1]) / ((1.0 - sy) / 1.0) * ((y - sy) / y);
    }
    if (fi > 1) sum += max(0.0, Time(x, y, L[fi - 1], R[fi - 1]));
    if (se < n && se + 1 != fi - 1)
      sum += max(0.0, Time(x, y, L[se + 1], R[se + 1]));
    printf("%.20f\n", sum);
  }
}